Chapter 3 Derivatives. Can u imagine if, anything on that bridge wasn’t move?
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Transcript of Chapter 3 Derivatives. Can u imagine if, anything on that bridge wasn’t move?
Why Do We Need Derivatives?
• In life, things are constantly “changing”• Such changes are described
mathematically by derivatives. • A “derivative” describes how something is
changing with respect to something
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Getting Started
Let’s start simple. Let’s start simple. •Consider the function Consider the function y(x)y(x) = 3 = 3•How does y change as a function of x?How does y change as a function of x?•It doesn’t change. It is a constant value of 3 everywhere.It doesn’t change. It is a constant value of 3 everywhere.””•how would we describe your response mathematically? how would we describe your response mathematically?
y(x) = 3
x1x2
Well, we define the change in the function y(x) with respect to the variable x, Δy/Δx, to be
then
Therefore, the function y(x) does not change over the interval x1 and x2
This mathematics will be how we describe the change in any function with respect to a given variable.
Mathematics of “Change”
.)()(
12
12
xxxyxY
xy
.033
12
xxxy
The symbol “Δ,” called delta, represents the change in a variable. Translated, Δy reads: “The change in y,” which equals:“y final minus y initial” or mathematically: “Δy = yfinal – yinitial.”
Understanding the symbols…
Example of a Straight Line
Consider the function y(x) = 3x + 1how would y(x) change with x? Let’s look at the interval between x = 2 and x = 3:
3 1
710
23
]1)2(*3[]1)3(*3[
23
)2()3(
xyxy
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y
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Y
y(x) = 3x + 1
Δy=3
Δx=1
Δy/Δx = 3The function y changes by 3 units between x=2 and x=3.
The slope of the line equals Δy/Δx
More than Straight Lines
x
yHow would you describe how the function y(x) changes at any given value of x?
From what you just learned, the change in the function y(x) with respect to x should be given by the slope of a straight line. But what is the straight line???
Graphs that CurveLet choose a point X on the graph“How does y(x) change at the point X?
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the change in the change in y(x)y(x) with respect with respect to to xx at point at point XX is given by the is given by the slope of the slope of the tangent linetangent line at at point point XX !!! !!!
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Slope of the tangent line at point X gives the change in y(x) with respect to x.:Slope=Δy/Δx.
The question now is “How do we determine the slope of the tangent line at point X?
Understanding the terminology…The word “tangent line” describes a line that intersects or touches a curve at only one point. At any given point on a smooth curve, there is only one unique tangent line and therefore there is only one value for the slope of that tangent line at that point.
Determining the Slope of the Tangent Line
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y
X X+h
the slope of the line will be Δy/Δx:
h
y(X)h)y(X
Xh)(X
y(X)h)y(XΔx
Δy
This line is still not the tangent line at point X, but we can make it look more like the tangent line if we make the h a smaller value:
X+h
Δy=y(X+h)–y(X)
Δx=(X+h)–X=h
Because h is smaller, the point X+h is closer to the original point X than before. Therefore, the slope of this second line is closer in value to the slope of the tangent line than what the slope of the first line was.
So What is a Derivative Anyway?
the line that passes through the points X and X+h will start looking more and more like the line tangent to the curve at point X as h goes to zero, then the line that goes through X and X+h will become the tangent line!!!
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Slope of tangent line = )zerotogoesas()()(
hh
XyhXy
dx
dy
X
dx
dywe call the derivative of y with respect to x.
the term “derivative” represents how the function y(x) instantaneously changes with respect to the variable x. As h goes to zero, Δy/Δx becomes dy/dx.
X+hX+hX+h
Tangent Line
So, let’s write this out in mathematics…h gets smaller
h gets even smaller
h goes to zero!
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Using the Definition for Derivatives
Let us now apply our newly derived formula to calculate the derivative of y(x) = x2.
x
hhx
hh
hxh
hh
xhxhx
hh
xhx
hh
xyhxy
dx
dy
2
)zerotogoesas(2
)zerotogoesas(2
)zerotogoesas()2(
)zerotogoesas()())((
)zerotogoesas()()(
2
222
22
xdx
dy2
Therefore, the slope of the tangent line that passes through x = 5 has a slope of 10!
Tangent line at x = 5 has a slope of 10. Therefore, the function y(x) has an instantaneous slope of 10 units at x=5.
y = x2
Graphing the DerivativeIf we have function y = x2
The derivative to be a function itself: dy/dx = 2x
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y
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-5
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x
dy/dx
Let us see how the two graphs are related.
• You know that the derivative of a function is just the slope of that function. For example, look at the graph of y = x2, for negative values of x, the slope of the tangent line should be negative. Looking at the graph of dy/dx, when x is negative, dy/dx is also negative!
The slope of the tangent line at x = -9 is -18.
-18
-8
The slope of the tangent line at x = -4 is -8.
y = x2
dy/dx = 2x
• When the slope of the tangent line equals zero, then the value of the derivative will equal zero!
The slope of the tangent line at x = 0 is 0.
Another Example
Suppose y =4x3-15x2+20
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y=4x3–15x2+20
xxdx
dy
hhxhxhx
hh
hxhhxhhx
hh
xxhxhxhxhhxx
hh
xxhxhx
hh
xyhxy
dx
dy
3012
)zerotogoesas(15301212
)zerotogoesas(15301212
)zerotogoesas(]20154[]20)2(15)33(4[
)zerotogoesas(]20154[]20)(15)(4[
)zerotogoesas()()(
2
22
2322
23223223
2323
Definition of derivative
Substituted in the expression for y(x)
Terms that survived after some terms canceled
Divided each term by h
These terms survived after h went to zero
Example Continued
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y=4x3–15x2+20
-40
-20
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-3 -2 -1 0 1 2 3 4 5
x
dy/dx
dy/dx = 12x2-30x
• The original function y(x) in the region between x = -2 and x = 0 should have a positive slope.
In this region, dy/dx should be positive.
Indeed, dy/dx has positive values between x=-2 and x=0.
• At x = 0 and at x = 2.5, y(x) has critical points (points where the slope of the tangent line equals zero) and therefore its derivative should equal zero at those points.
The slope of the tangent lines should be zero at these points.
Indeed, at x = 0 and x = 2.5, dy/dx equals zero.
• Between x = 0 and 2.5, y(x) is decreasing in value which implies that its derivative is negative in this region.
Between x = 0 and x = 2.5, y(x) should have a negative slope.
Indeed, between x=0 and x=2.5, dy/dx is negative in value.
The Shortcut…Certainly, the definition for the derivative can be used each time when one needs to be determined, but there exists a shortcut when it comes to functions of the form: y(x) = Axn, where “A” is just a numerical constant and “n” is an integer, positive or negative.
dy/dx = nAxn-1
Using this shortcut to calculate the derivative of y(x) = x2, we get:dy/dx = 2*x2-1 = 2x.
In our second example, we found that the derivative of y =4x3-15x2+20 to be:dy/dx = 12x2 – 30x
Some Simple Rules of Differentiation
What are these symbols?
“f” and “g” are functions of x: f(x) and g(x).
“c” represents a constant numerical value and therefore is not a function of x.
“n” represents an integer number, positive or negative.f’ is shorthand for df/dx. Likewise, g’ is shorthand for dg/dx.
0][)1 cdxd
''][)5 gfgfdxd
'*'*]*[)7 fggfgfdxd
1][)3 nn ncxcxdxd
'][)4 cfcfdxd
xxdxd
cos][sin)8
1][)2 xdxd
'][)6 1fnffdxd nn
xxdxd 2sec][tan)10 xx
dxd
sin][cos)9